1. **State the problem:** We have a right triangle AOB with right angle at O, OA = 12 cm, OB = 5 cm. We need to find angle BAO and the area of the shaded region between two arcs centered at A and B. 2. **Find angle BAO:** Since triangle AOB is right angled at O, by Pythagoras, AB is the hypotenuse: $$AB = \sqrt{OA^2 + OB^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.$$ 3. Angle BAO is the angle at A between BA and AO. Using cosine rule or trigonometry: $$\cos(\angle BAO) = \frac{OA}{AB} = \frac{12}{13}.$$ 4. Calculate angle BAO: $$\angle BAO = \cos^{-1}\left(\frac{12}{13}\right) \approx 0.3948 \text{ radians}.$$ 5. **Calculate the area of the shaded region:** The shaded region lies between two arcs: - Arc OP with center A and radius OA = 12 cm - Arc OQ with center B and radius OB = 5 cm 6. Length of AB is 13 cm, so points P and Q lie on AB. The shaded region is the area between the two circular sectors formed by arcs OP and OQ minus the triangle OAB. 7. Area of sector with center A and radius 12, angle BAO = 0.3948 radians: $$\text{Area}_A = \frac{1}{2} \times 12^2 \times 0.3948 = \frac{1}{2} \times 144 \times 0.3948 = 28.4256 \text{ cm}^2.$$ 8. Area of sector with center B and radius 5, angle ABO (angle at B) is: Since triangle is right angled at O, angle ABO = \angle BAO complement to right angle: $$\angle ABO = \frac{\pi}{2} - 0.3948 = 1.1750 \text{ radians}.$$ 9. Area of sector with center B: $$\text{Area}_B = \frac{1}{2} \times 5^2 \times 1.1750 = \frac{1}{2} \times 25 \times 1.1750 = 14.6875 \text{ cm}^2.$$ 10. Area of triangle AOB: $$\text{Area}_{\triangle} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2.$$ 11. The shaded area is the sum of the two sectors minus the triangle area: $$\text{Shaded area} = 28.4256 + 14.6875 - 30 = 13.1131 \text{ cm}^2.$$ **Final answers:** - (i) $\angle BAO = 0.3948$ radians - (ii) Area of shaded region $= 13.1131$ cm$^2$